Can non-constant polynomials over algebraically closed fields never vanish? No, right?
Thoughts: Take $f(x,y)$, suppose that for whatever $y'$ we plug in for $y$ we find that $f(x,y')=c$, that is it kills the $x$ term. This would mean that $f(x,y)$ is really just a polynomial in $y$ and boom we are done, because a polynomial in one variable has a root in a a.c. field. Suppose that's not the case, i.e. if we plug in some $y'$ we just get a polynomial in $x$, then by same reasoning, the a.c. guarantees a root. Same argument applies to more variables.
So the main task is: given two affine varieties that are disjoint, $V$ and $V'$, construct a regular function that vanishes on $V$ and is constant $c$ on $V'$. In the simplest case we just have that each of these varieties is defined by a single polynomial each, $f$ and $g$, that differ by a constant, such that $f-g=d$. Then the polynomial we want would be $cf/(f-g)$, because for a point on $V$ we know $f=0$ so this gives $0/(-g)$ (and we know $g\neq0$ on $V$ since $V\cap V'= \emptyset$) and for a point on $V'$ we know $g=0$ which gives $cf/f$ (and we know $f\neq0$ on $V'$ since $V\cap V'= \emptyset$) so this is just $c$. However, I am not sure how to generalize this to varieties defined by polynomials that aren't just different by additive constants and then generalize it more so to varieties defined by multiple polynomials.
I mean, can we deduce from the fact if $f-g$ is a polynomial (rather than a constant) that the varieties $f$ and $g$ define intersect? If $f-g$ is a polynomial, then it must vanish thus there are points where $f=g$. However, these points don't necessarily need to be points where $f=g=0$, and thus I don't think the varieties need to intersect necessarily, but I'm not sure.